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Ludmilka [50]
3 years ago
11

Explain how the gas laws apply to the act of breathing. Describe the changes that occur in each step of the process in terms of

the pressure, volume, temperature, and moles of gas present for each step, identify which variables are constant, and which are changing name the gas law or gas laws that apply to the process create and solve a sample calculation using realistic values of pressure, temperature, amount of gas, and volume. Please HELP!
Physics
1 answer:
MrRa [10]3 years ago
4 0
<span>When you in inhale, you move your diaphragm down increasing the volume of the cavity in your ribs (its called pleural cavity). This lowers the pressure inside your lungs. When the pressure in your lungs is less than the atmospheric pressure, air draws inwards. Ideal gas law: PV=nRT -> assume nRT are constant for our purposes. then we allow one system to be equal to the same system at different conditions, thus: P1V1=P2V2 (Boyles Law) assume P1=1 atm and V1=1 L (keeping the numbers simple to avoid confusion). when the V is increased (for example say doubled) V2=2 L, then: (1 atm)*(1 L)=P2*(2 L) -> P2=0.5 atm The pressure was decreased. Apply the same logic for exhalation.</span>
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Answer:

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A coil is wrapped with 300 turns of wire on the perimeter of a circular frame (radius = 8.0 cm). Each turn has the same area, eq
TiliK225 [7]

Answer:

Approximately 18 volts when the magnetic field strength increases from \rm 20\; mT to \rm 80\;mT at a constant rate.

Explanation:

By the Faraday's Law of Induction, the EMF \epsilon that a changing magnetic flux induces in a coil is:

\displaystyle \epsilon = N \cdot \frac{d\phi}{dt},

where

  • N is the number of turns in the coil, and
  • \displaystyle \frac{d\phi}{dt} is the rate of change in magnetic flux through this coil.

However, for a coil the magnetic flux \phi is equal to

\phi = B \cdot A\cdot \cos{\theta},

where

  • B is the magnetic field strength at the coil, and
  • A\cdot \cos{\theta} is the area of the coil perpendicular to the magnetic field.

For this coil, the magnetic field is perpendicular to coil, so \theta = 0 and A\cdot \cos{\theta} = A. The area of this circular coil is equal to \pi\cdot r^{2} = \pi\times 8.0\times 10^{-2}\approx \rm 0.0201062\; m^{2}.

A\cdot \cos{\theta} = A doesn't change, so the rate of change in the magnetic flux \phi through the coil depends only on the rate of change in the magnetic field strength B. The size of the magnetic field at the instant that B = \rm 50\; mT will not matter as long as the rate of change in B is constant.

\displaystyle \begin{aligned} \frac{d\phi}{dt} &= \frac{\Delta B}{\Delta t}\times A \\&= \rm \frac{80\times 10^{-3}\; T- 20\times 10^{-3}\; T}{20\times 10^{-3}\; s}\times 0.0201062\;m^{2}\\&= \rm 0.0603186\; T\cdot m^{2}\cdot s^{-1}\end{aligned}.

As a result,

\displaystyle \epsilon = N \cdot \frac{d\phi}{dt} = \rm 300 \times 0.0603186\; T\cdot m^{2}\cdot s^{-1} \approx 18\; V.

6 0
3 years ago
I need help on a homework question.
Mama L [17]

Answer:

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7 0
3 years ago
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The diagram shows the electric field around two charged objects.
anyanavicka [17]

Here we have been given two charged body W and X.

We are asked to determine the nature of charge.

Before coming into conclusion, first we have to understand the electric field lines.

The electric field lines are pictorial representation of imaginary lines which are drawn to denote the electric field in a graphical  way.

The electric field line starts from a positive charge and ends at a negative charge.It means for positive charges the electric field lines are outward and for negative charges the filed lines will be inward.

In the given diagram,the filed lines for W is towards W itself.The same is also in case of X.

Hence both the charges must be negative in nature.

Hence the correct answer to the question will be B i.e W negative and X: NEGATIVE.

8 0
3 years ago
Read 2 more answers
A 11.3-kg object oscillates at the end of a vertical spring that has a spring constant of 2.20 ✕ 104 N/m. The effect of air resi
3241004551 [841]

Answer:

(a) the frequency of the dampened oscillation is 7.02 Hz

(b) percentage decrease in amplitude of the oscillation in each cycle is 2%

Explanation:

Given;

mass of the object = 11.3 kg

the spring constant = 2.2 X 10⁴ N/m.

damping coefficient b = 3.00 N · s/m

Part (a) the frequency of the dampened oscillation

The oscillation frequency is calculated as follows;

\omega _D = \sqrt{\omega_o^2 -(\frac{b}{2m})^2}\\\\\omega_o^2 = \frac{k}{m} =\frac{2.2X10^{4}}{11.3} = 1946.903rad/s\\\\thus, \omega _D = \sqrt{1946.903-(\frac{3}{2*11.3})^2} =44.12 rad/s

The damped frequency = \frac{\omega _D}{2\pi } =  \frac{44.12}{2\pi } = 7.02 Hz

Part (b)  percentage decrease in amplitude of the oscillation in each cycle

The amplitude of the oscillation depends on the damping coefficient (b) and period (T), and it is given as;

A(t) = e^{-\frac{b}{2m}(t)}

After one cycle, the amplitude changes from A(t) to A(t+T), where T is the period of the oscillation.

A(t +T) = e^{-\frac{b}{2m}(t+T)}

Percentage decrease in amplitude is gotten by dividing A(t) by A(t+T)

= \frac{e^{-\frac{b}{2m}(t)}}{e^{-\frac{b}{2m}(t+T)}} =e^{-\frac{b}{2m}(T)}

But T = 1/f

Substituting the values of the parameters in the above equation, we will have;

=e^{-\frac{b}{2m}(T)} = e^{-\frac{3}{2X11.3}(\frac{1}{7.02})} = 0.98

Percentage decrease = 1 - 0.98 = 0.02 = 2%

4 0
3 years ago
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